The projection of your solid down to the $xy$-plane is described by
$$5(x^2+y^2) \leq 6-7x^2-y^2,$$
which simplifies to
$$12x^2+6y^2 \leq 6.$$
This is an equation of a solid ellipse $E$ compressed by the factor $\sqrt{2}$ in the $x$-direction. The volume can be expressed as
$$\iint\limits_E (6-(12x^2+6y^2)) \, dA.$$

A truly efficient way to evaluate this integral is to work in an elliptic coordinate system:
$$x=r\cos(t)/\sqrt{2}, \; y=r\sin(t).$$ In this coordinate system, the function simplifies to $6-6r^2$ and the area element is $dA=(r/\sqrt{2})\, dr \, dt$. The area element is similar to the familiar $dA = r\,dr\,d\theta$ in polar coordinates, but accounts for the extra compression in the $x$-direction. The formula can also be derived using the Jacobian.

Using this coordinate system, the integral can now be expressed as
$$\int_0^{2\pi} \int_0^1 (6-6r^2)r/\sqrt{2} \, dr\, dt = 3\pi/\sqrt{2},$$
which is quite easy to do by hand.

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